pub fn new_birthday_probability(n: u32) -> f64 {
    // 1-至多一人在该日过生日的概率
    let mut p: f64 = 1.0;
    if n > 365 {
        return 0.0;
    }
    for i in 0..n {
        p = p * (365 - i) as f64 / 365.0;
    }
    p = 1.0 - p;
    p
}
